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An order for 160 of a product is processed on work centers A and B. The setup time on A is 30 minutes, and run time is ten mi
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Answer #1

For the lot of 160 units :

Time taken by operation A =Set up time +run time

=30+160*10=1630

Time taken to move lot from A to B=10

Time taken by operation B =Set up time +run time

=40+160*6=1000

Total time =1630+10+1000=2640 minutes

The correct answer is option C.None of them

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In this case ,we split the lot into L1 and L2 such that the idle time at B is minimum or zero for the second lot L2.

Also since there is no queue at either workstations ,

Time taken by operation A for lot L2 + 10 >=Time taken by operation B for Lot L1

The difference between these times should be minimum or zero.

B9 Set up time Run time L1 time L2 time 970 485 Idle time 3 Operation A 4 Operation B 5 Total 30 1000 540 265 805 50 535 2350

The shortest total manufacturing lead time occurs at splitting lot into L1:97 and L2:53 with an idle time of 5 minutes.

Hence,the correct answer is option E.None of them

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